Question

For 0<x<5 and t>0 , solve the one dimensional heat flow equation Thita u / Thita t = 4 thita^2 u / Thita x^2 satisfying the condition u(t,0) = u(t,5) = 0 , u(0,x) = x

Answer 1

We use the method of separation of variables

$$u(x,t)=X(x)T(t)\to\left\{\begin{array}{l} \displaystyle\frac{\partial u}{\partial t}=\displaystyle\frac{\partial\left(X(x)T(t)\right)}{\partial t}=X(x)T'(t)\\[0.3cm] \displaystyle\frac{\partial^2 u}{\partial x^2}=\displaystyle\frac{\partial^2\left(X(x)T(t)\right)}{\partial x^2}=X''(x)T(t) \end{array}\right.$$

Then,

$$\frac{\partial u}{\partial t}=4\cdot\frac{\partial^2 u}{\partial x^2}\to\left.X(x)T'(t)=4X''(x)T(t)\right|\div\left(4X(x)T(t)\right)\\[0.3cm] \frac{T'(t)}{4T(t)}=\frac{X''(x)}{X(x)}=-\lambda\to \left\{\begin{array}{l} X''(x)+\lambda X(x)=0\\[0.3cm] T'(t)+4\lambda T(t)=0 \end{array}\right.$$

The boundary conditions take the form

$$\left\{\begin{array}{l} u(0,t)=X(0)T(t)=0\to\boxed{X(0)=0}\\[0.3cm] u(5,t)=X(5)T(t)=0\to\boxed{X(5)=0} \end{array}\right.$$

First, we solve the Sturm-Liouville problem for the function $X(x)$

In our case,

$$\left\{\begin{array}{l} X''(x)+\lambda X(x)=0\\[0.1cm] X(0)=0\\[0.1cm] X(5)=0 \end{array}\right.$$

We are looking for a solution in the form

$$X(x)=e^{kx}\to X''(x)=k^2\cdot e^{kx}\to\\[0.3cm] k^2\cdot e^{kx}+\lambda\cdot e^{kx}=0\to e^{kx}\cdot\left(k^2+\lambda\right)=0\to k^2=-\lambda\\[0.3cm] \left[\begin{array}{l} k_1=\sqrt{-\lambda}=i\sqrt{\lambda}\\[0.1cm] k_2=-\sqrt{-\lambda}=-i\sqrt{\lambda} \end{array}\right.$$

Conclusion,

$$X(x)=A_1e^{i\sqrt{\lambda}x}+A_2e^{-i\sqrt{\lambda}x}\equiv C_1\cos\left(\sqrt{\lambda}x\right)+C_2\sin\left(\sqrt{\lambda}x\right)\\[0.1cm] X(0)=0=C_1\cos\left(\sqrt{\lambda}0\right)+C_2\sin\left(\sqrt{\lambda}0\right)\\[0.1cm] \boxed{C_1=0}\\[0.1cm] X(5)=0=C_2\sin\left(5\sqrt{\lambda}\right)\to5\sqrt{\lambda}=n\pi,n\in\mathbb{N}\\[0.1cm] \boxed{\lambda_n=\left(\frac{n\pi}{5}\right)^2,n\in\mathbb{N}}$$

Then,

$$\boxed{X_n(x)=C_n\cdot\sin\left(\frac{n\pi x}{5}\right),n\in\mathbb{N}}$$

For the function $T(t)$ it is necessary to solve the equation

$$T'(t)+\lambda_nT(t)=0\to\frac{dT}{dt}=-\lambda T(t)\to\frac{dT}{T}=-\lambda_n dt\\[0.1cm] \ln|T(t)|=-\lambda_nt+\ln|A_n|\to\boxed{T_n(t)=A_n\cdot e^{-\lambda_n t}}$$

Then, a particular solution to the equation has the form

$$u_n(x,t)=X_n(x)T_n(t)=A_n\cdot e^{-\lambda_nt}\cdot C_n\cdot\sin\left(\frac{n\pi x}{5}\right)$$

And the general solution of the equation has the form

$$\boxed{u(x,t)=\sum\limits_{n=1}^{\infty}u_n(x,t)=\sum\limits_{n=1}^{\infty}B_ne^{-\lambda_nt}\sin\left(\frac{n\pi x}{5}\right)}$$

It remains to find an expression for the coefficients $B_n$. To do this, we use the initial condition, as well as the condition of orthogonality of the function of the Sturm-Liouville problem (without proof)

$$\int\limits_0^5\sin\left(\frac{n\pi x}{5}\right)\cdot\sin\left(\frac{n\pi x}{5}\right)dx=\frac{5}{2}\\[0.3cm] \int\limits_0^5\sin\left(\frac{n\pi x}{5}\right)\cdot\sin\left(\frac{m\pi x}{5}\right)dx=0\\[0.3cm] \boxed{\int\limits_0^5\sin\left(\frac{n\pi x}{5}\right)\cdot\sin\left(\frac{m\pi x}{5}\right)dx=\frac{5}{2}\delta_{nm}}$$

Then,

$$u(x,0)=x=\sum\limits_{n=1}^{\infty}B_n\sin\left(\frac{n\pi x}{5}\right)\to\\[0.3cm] \int\limits_0^5\times\left|\sum\limits_{n=1}^{\infty}B_n\sin\left(\frac{n\pi x}{5}\right)=x\right|\times\sin\left(\frac{m\pi x}{5}\right)dx\to\\[0.3cm] \sum\limits_{n=1}^{\infty}B_n\int\limits_0^5\sin\left(\frac{n\pi x}{5}\right)\sin\left(\frac{m\pi x}{5}\right)dx=\int\limits_0^5 x\sin\left(\frac{m\pi x}{5}\right)dx\to\\[0.3cm] \sum\limits_{n=1}^{\infty}B_n\delta_{nm}\frac{5}{2}=\int\limits_0^5 x\sin\left(\frac{m\pi x}{5}\right)dx\to\\[0.3cm] B_m\frac{5}{2}=\int\limits_0^5 x\sin\left(\frac{m\pi x}{5}\right)dx$$

It remains to calculate the indicated integral. To do this, we use the method of integration by parts

$$\int\limits_0^5 x\sin\left(\frac{m\pi x}{5}\right)dx=\\[0.3cm] =\left.\frac{-5x}{m\pi}\cdot\cos\left(\frac{m\pi x}{5}\right)\right|_0^5-\int\limits_0^5\left(-\frac{5}{m\pi}\cos\left(\frac{m\pi x}{5}\right)\right)dx=\\[0.3cm] =-\frac{25}{m\pi}\cdot\cos(m\pi)+\left.\frac{25}{m^2\pi^2}\cdot\sin\left(\frac{m\pi x}{5}\right)\right|_0^5\to\\[0.3cm] \boxed{\int\limits_0^5 x\sin\left(\frac{m\pi x}{5}\right)dx=-\frac{25\cdot(-1)^m}{m\pi}}$$

Conclusion,

$$B_m\frac{5}{2}=\frac{25\cdot(-1)^{m+1}}{m\pi}\to\boxed{B_m=\frac{10\cdot(-1)^{m+1}}{m\pi}}$$

ANSWER

$$u(x,t)=\sum\limits_{n=1}^{\infty}\left(\frac{10\cdot(-1)^{n+1}}{n\pi}\right)e^{-\lambda_nt}\sin\left(\frac{n\pi x}{5}\right)$$

$$\lambda_n=\left(\frac{n\pi}{5}\right)^2,n\in\mathbb{N}$$